anonymous
  • anonymous
i dont understand this problem: 8^(x/2)=4^(x+1) anyone know?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i ASSume you want to solve for x?
anonymous
  • anonymous
i guess, the questions says to solve
anonymous
  • anonymous
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anonymous
  • anonymous
right - this equation has x in the exponent (power) so we take logs of both sides: ln 8^(x/2) = ln 4^(x+1) by the laws of logarithms: (x/2) ln 8 = (x+1) ln 4 (x/2) = ((x+1) ln4) / ln 8) (x/2) / (x+1) = ln4) / ln 8) (x/2) / (x+1) = 2/3 excuse me - i must leave the room - be back later to complete it
anonymous
  • anonymous
no problem. thanks for your reply in the meantime, jimmyrep
anonymous
  • anonymous
so whats left is to solve this last equation 3x = 4(x+1) 3x = 4x + 4 x = -4
anonymous
  • anonymous
the 3 main laws of logarithms are log x + log y = log xy log x - log y = log (x/y) log (x^y) = y log x
anonymous
  • anonymous
thanks jimmyrep

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